I'm trying to solve this problem from Hungerford V.5.9.
I have to show $x^{2^n}+x+1$ is irreducible over $\mathbb{Z}_2$ if n>2.
I would appreciate some hint cause I don't know how to start with it.Thanks!
EDITED: This question is completely different to On irreducible factors of $x^{2^n}+x+1$ in $\mathbb Z_2[x]$ The reason is obvious, i m asking if this polynomial is irreducible. In the other question, we assume it is reducible without showing it.
The claim is not true. Assume $n>1$ is odd. Then
$$ \begin{align} x^{2^n}+x+1 &= (x^{2^n}+x^{2^{n-1}}+1)+(x^{2^{n-1}}+x^{2^{n-2}}+1)+\cdots+(x^2+x+1)\\ &=\sum_{k=0}^{n-1}(x^{2^{k+1}}+x^{2^k}+1) = \sum_{k=0}^{n-1} (x^2+x+1)^{2^k}\\ &= (x^2+x+1)\left(\sum_{k=0}^{n-1} (x^2+x+1)^{2^k-1}\right) \end{align} $$
We used that $n$ is odd in the first equality, where there are $n$ ones on the right side, thus summing to $1$.
Note: The same argument can be used to show that $x^{2^n}+x+1$ is reducible whenever $n$ has an odd factor greater than $1$. Therefore it can only be irreducible if $n=2^m$.